Question

Practice question — Given the trapezoid to the right find the length of legs in the following isosceles trapezoid

Answer 1

Given:

A figure of an isosceles trapezoid with bases 18 and 24, and the vertical height is 4.

To find:

The legs of the isosceles trapezoid.

Solution:

Draw another perpendicular and name the vertices as shown in the below figure.

From the figure it is clear that the AEFD is a rectangle. So,

$EF=AD=18$

Since ABCD is an isosceles trapezoid, therefore in triangle ABE and DCF,

$AB=DC$               (Legs of isosceles trapezoid)

$AE=DF$                (Vertical height of isosceles trapezoid)

$m\angle AEB=m\angle DFC$             (Right angle)

$\Delta ABE\cong \Delta DCF$                (HL postulate)

$BE=CF$                (CPCTC)

Now,

$BE+EF+FC=BC$

$2BE+18=24$

$2BE=24-18$

$2BE=6$

$BE=3$

Using Pythagoras theorem in triangle ABE, we get

$Hypotenuse^2=Perpendicular^2+Base^2$

$(AB)^2=(AE)^2+(BE)^2$

$(AB)^2=(4)^2+(3)^2$

$(AB)^2=16+9$

$(AB)^2=25$

Taking square root on both sides, we get

$AB=\pm \sqrt{25}$

$AB=\pm 5$

Side length cannot be negative. So, $AB=5$.

Therefore, the length of legs in the given isosceles trapezoid is 5 units.